206 XIII INTEGRATION
if M0 n N0 is p-negligible. Putting M = M0 - (M0 n N0) and N =
NO - (M0 n NO), this condition is equivalent to the relations g = q>Mg and
h = <pN/z almost everywhere with respect to p (13.15.3). But g = <pMg almost
everywhere with respect to p if and only if p = cpM • n (13.15.3) and (13.14.5)),
or equivalently if and only if \L is concentrated on M. This proves the first
assertion. The second follows by replacing M and N by universally measurable
sets M' G M and N' c N such that |/i|(M - M7) = 0 and |v|(N - N') = 0
In particular, if ju is any real measure, then u+ and u~ are concentrated on
two disjoint ^-measurable sets, each of which is both u+-measurable and
/T-measurable ((13.15.3) and (13.16.1)).
'(13.18.2) (i) Ifk is a measure disjoint from each of two measures ju, v, then X
is disjoint from ju + v.
(ii) IfH is a set of positive measures disjoint from a measure v, and //H is
bounded above, then \JL = sup H (13.15.4) is disjoint from v.
(i) We may assume that X, u, v are all positive and of the forms 1 =/• p,
jj. = g ' p, v = h * p, where p is a positive measure and/, g, h are positive. The
result then follows from the inequality inf(/, g + h) ^ inf(/, g] + inf(/, h).
(ii) We may assume that v ^ 0 and that H consists of an increasing
sequence (/O (13.15.4), and we may write v = h - p, un = gn - p, p = (sup gn\ - p,
\ » /
where p is a positive measure and/and the gn are positive functions. Then the
result follows from the formula inf//, sup gn\ = sup /inf(/, gn)\.
(13.18.3) If two measures fi, v are disjoint, then \fj, + v| = \JJL\ -h |v|.
Write ju = g • p and v = h - p, where p is a positive measure and
inf(|#|, \h\) = 0; then use the relation \g 4- h\ = \g\ + \h\.
(13.18.4) (Lebesgue's decomposition theorem) Let \JL be a positive meas-
ure on X. Then every complex measure v on X is uniquely expressible in the
form v = v' + v", where v' has base n and v" is disjoint from u. Ifv*tQ, then v'
andv" are positive, and V = sup(inf(v, nu)).
If vi + vj = v'2 + V2, where vj and v^ are disjoint from /z, and vi and v'2
are measures with base /z, then we have v'[ — v£ = v^ — vj, and since v'2 - v[ is
disjoint from p (13.18.2), it can be a measure with base u only if it is zero, by
virtue of (13.18.1). This establishes the uniqueness of the decomposition. Forligible if and onlygmen-Lindelof principle as in Section 9.5, Problem 17, by taking g(s) = e*2.)